Last Updated on September 12, 2026 by Maged kamel
Nominal Shear strength for M10x7.50-use CM-15.
Compute the Nominal Shear strength for M10x7.50.
We have a given M10x7.50 steel section of A572 Grade 65; we need to compute the nominal Shear strength. We will apply CM#15 and the related specification 2016 in this example.
First, we will use Table 2-4 to find the yield stress Fy value for A572 steel grade 65. This steel has a yield stress Fy = 6565 ksi and an ultimate stress Fu = 80 ksi.
The next question concerns the table used to obtain the complete data for M10x7.5; in Part 1, we will find that Tables 1-2 are used to get the section information.
The data for M10x7.50 comes from Tables 1-2 in Part 1. The Flange width bf is 2.69 inches, and the thickness is 0.173 inches.
The depth is d = 9.99 inches, and the web thickness is 0.13 inches. The most critical parameter is h/tw, which we will find in part 2 of the same table.
Find h/tw from Table 1-3, part 2.
The ratio h/tw equals 71. h is the distance between the web’s two filleted portions. The following slide image shows part 2 of Table 1-2.
Find the limiting h/w based on Fy.
The limiting h/tw and the yield stress are related by the equation h/tw = 2.24*sqrt (E/Fy), as shown in the next slide image. Based on the yield stress Fy of the given M10x7.5 for grade 65, the limiting h/w equals 2.24*sqrt (29000/65)= 47.30.
The following slide image shows the relation of h/tw and the nominal Shear value. There are three zones; the first Zone is from zero to 47.30, marked as point 1 on the Zonexis, obtained from the relation 2.24*sqrt(E/Fy); in this Zone, the term used is Cv1, unlike CV in CM#14, which is Zone 1, and φv equals 1.
The equation gives the nominal Shear value for nominal Shear Vn = Cv*0.60Fy*d*tw and can be written as 1*0.60*Fy*Aw.
The second Zone is from h/tw equals 47.30 to 53.692, marked as Zone2 on the x-axis, and is obtained from the relation 1.1*sqrt(kv *E/Fy); in this Zone, Cv equals 1, and φv equals 0.90. The Kv fafactor is Zone.34 in spSpecification016. The nominal Shear value is Vn = Cv*0.60Fy*d*tw, nd it can be written as 1*0.60Fy*Aw.
The third Zone is from h/tw, which equals 53.69, to the end; Cv zones less than 1 in this Zone, and φv equals 0.90. The nominal Shear value is determined by the equation Vn=Cv1*0.60Fy*d*tw; Cv1 equals 53.69/h/tw. For our section M10x7.5 grade 65, h/tw=71.0, and the value of cv1=53.69/71=0.756. Please refer to the next slide image.
The nominal Shear strength for M10x7.50.
The area of the web for M10x7.50 equals the product of d*tw, or (9.99*0.13), or 1.2987 inch2; the cv1 value equals 0.7562, and the nominal shear equals 0.7562*(0.60*Fy)*(Aw)=0.7562*0.6*65*1.2987=38.30 kips.
Excel plot for h/tw versus Vn.
I used an Excel plot of the relationship between h/tw and the nominal Shear Vn, which shows the Shear zones. Zone 1 starts at an h/tw value from 0 to 47.30, and the nominal Shear value equals 50.65 Kips
. Zone 2 begins for h/tw from 47.30 to 53.69, and the nominal Shear value equals 50.65 Kips. Zone 3 starts from h/tw greater than 53.69; The Nominal Shear Vn equals 50.41 kips for h/tw equals 71; the section has a nominal Shear Vn value of 38.30 kips for h/tw=71.0. Thanks a lot.
You can view or download the PDF used to solve this problem from the following link.
Here is the link to Chapter 8 – Bending Members, section A, Beginner’s Guide to the Steel Construction Manual, 14th ed.
Here is the link to Chapter 8 – Bending Members, Section A, Beginner’s Guide to the Steel Construction Manual, 15th ed.
Here is the link to Chapter 8 – Bending Members, Section A, Beginner’s Guide to the Steel Construction Manual, 16th ed.
The previous post is on nominal Shear strength for S41x121.